Explanation Given info. a → × ( b → × c → ) = ( a → . c → ) b → − ( a → . c → ) b → N → × B → = T → B → × T → = N → Concept used: B → = T → × N → Calculation: N → × B → ( putting the value of B → = T → × N → ) = N → × ( T → × N → ) ( used: a → × ( b → × c → ) = ( a → . c → ) b → − ( a → . b → ) c → ) = ( N → . N → ) T → − ( N → . T → ) N → ( ( N → . N → = | N → | | N → | cos 0 = 1 ) , N → ...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Sequence and Series

A sequence is defined as a collection of things. Series is defined to sum the things one by one in the sequence. It was invented by German mathematician Carl Friedrich Gauss.

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Chapter 14.4, Problem 89E

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Use the identity $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} . \vec{c}) \vec{b} - (\vec{a} . \vec{b}) \vec{c}$ to prove $\vec{N} \times \vec{B} = \vec{T} and \vec{B} \times \vec{T} = \vec{N}$

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Chapter 14.4, Problem 88E

Chapter 14.4, Problem 90E

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Let G = <a, b|a2 = b4 = e, ab = b3a> Express a3b2abab3 in the form biaj, where 0 <=i <=1 and 0<=j<=3.

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Chapter 14 Solutions

CALCULUS (CLOTH)

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 6PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E

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Use the identity a → × ( b → × c → ) = ( a → . c → ) b → − ( a → . b → ) c → to prove N → × B → = T → and B → × T → = N →

Solution Summary: The author explains how to determine the identity of the person.

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Use the identity $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} . \vec{c}) \vec{b} - (\vec{a} . \vec{b}) \vec{c}$ to prove $\vec{N} \times \vec{B} = \vec{T} and \vec{B} \times \vec{T} = \vec{N}$

Want to see the full answer?

Chapter 14 Solutions

Use the identity a → × ( b → × c → ) = ( a → . c → ) b → − ( a → . b → ) c → to prove N → × B → = T → and B → × T → = N →

Solution Summary: The author explains how to determine the identity of the person.

Concept explainers

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Use the identity a→×(b→×c→)=(a→.c→)b→−(a→.b→)c→ to prove N→×B→=T→ and B→×T→=N→

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Chapter 14 Solutions

Use the identity $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} . \vec{c}) \vec{b} - (\vec{a} . \vec{b}) \vec{c}$ to prove $\vec{N} \times \vec{B} = \vec{T} and \vec{B} \times \vec{T} = \vec{N}$